Let ABC be an equilateral triangle. Let P be a point on the side AC and Q be a point on the side AB so that both triangles ABP and ACQ are acute. Let R be the orthocentre of triangle ABP and S be the orthocentre of triangle ACQ. Let T be the point common to the segments BP and CQ. Find all possible values of ∠CBP and ∠BCQ such that triangle TRS is equilateral.

So for TR = TS one obvious solution is that α = β, ψ = σ to make the corresponding terms of TR˛ and TR˛ above equal, and when α = β the points P and Q are symmetrical across AI where I is the foot of A to BC.

Since SA = SB and CG ⊥ AB and CQ ⊥ AM, we then also have

∠ABS = ∠SAB = ∠TCS = σ

Assume a solution has been attained and that ∠CBP = α1 and ∠BCQ = β1

We will prove that for every unique value α1 for ∠CBP there is one and only one corresponding value β1 for ∠BCQ to satisfy the problem.

Indeed, let’s keep angle α1 and increase ∠BCQ. As we do so point T moves to T’ closer to N and RT’ < RT, or RT decreases.

We also know that ∠MAN = α + β. So ∠MAN increases by the same amount of the increase of ∠BCQ, and ∠SAG also simulatneously decreases by the same amount. Therefore, as we increase ∠BCQ, point S moves to S’ closer to point G and RS’ > RS, or RS increases.

The same but opposite effect occurs if we decrease ∠BCQ.

Therefore TR will no longer equal SR if ∠BCQ ≠ β1. So for every angle α there is only one unique angle β to satisfy the condition for triangle TRS to be equilateral.

We earlier proved that ∠BCQ = ∠CBP is a condition for ST = RT. Based on the above reasoning, there is no other value for ∠CBP to equal ∠BCQ except that their values are the same, or α = β and ψ = σ, and point T has to always be on AI. Now let’s find ∠α.

Since triangle TRS is equilateral, and R is on the bisector BH of ∠ABC and T is on bisector AI of ∠BAC, we have SR || BC, ST || AC and RT || AB, or BH is the bisector of ∠SBT or δ = ψ. We have